Complex number problem $i \tan (\theta)$ proof.

Question

Given $z = \cos (\theta) + i \sin (\theta)$, prove $\dfrac{z^{2}-1}{z^{2}+1} = i \tan(\theta)$

I know $|z|=1$ so its locus is a circle of radius $1$; and so $z^{2}$ is also on the circle but with argument $2\theta$; and $z^{2}+1$ has argument $\theta$ (isosceles triangle) so it lies on a line through the origin and $z$.

$z^{2}-1$, $z^{2}$, and $z^{2}-1$ all lie on a horizontal line $i \sin (\theta)$.

On the Argand diagram I can show $z^{2}+1$ and $z^{2}-1$ are perpendicular so the result follows.

Can anyone give an algebraic proof?

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Oct 15 '18 at 07:06
@badjohn: Argan (no d) was the first name of the main character in the famous "The Imaginary Invalid" by Molière. Armand is a little outmoded but still in use. ;-) — , Oct 15 '18 at 07:50
Joking? I only know Argan as an ingredient in some fancy shampoos. I know Armand as a name but, not as far as I know, an appropriate one here. Are you saying that Armand may be appropriate? Wolfram doesn't seem to know it: http://mathworld.wolfram.com/ArgandDiagram.html. — badjohn, Oct 15 '18 at 07:58

score 3 · Accepted Answer · answered Oct 15 '18 at 07:43

3

As $z\ne0,$ with $|z|=1$

$$\dfrac{z^2-1}{z^2+1}=\dfrac{z-\dfrac1z}{z+\dfrac1z}$$

Now $\dfrac1z=\dfrac1{\cos\theta+i\sin\theta}=\cos\theta-i\sin\theta$

answered Oct 15 '18 at 07:43

lab bhattacharjee

274,582

1

That is elegant! – Martin R Oct 15 '18 at 07:45
Brilliant. Thank you. – Rik Oct 15 '18 at 11:46

Martin R · Answer 2 · 2018-10-15T07:41:19.447

For $z = \cos \theta + i \sin \theta$ you have $$ \frac{z^2-1}{z^2+1} = \frac{ \cos^2 \theta - \sin^2 \theta + 2i \cos\theta \sin \theta -1}{\cos^2 \theta - \sin^2 \theta + 2i \cos\theta \sin \theta +1 } \, . $$ Now substitute $1 = \cos^2 \theta + \sin^2 \theta $ in both numerator and denominator, and collect terms: $$ \frac{-2 \sin^2 \theta + 2 i \cos\theta \sin\theta}{2 \cos^2\theta + 2 i \cos\theta \sin\theta} = \frac{- \sin\theta + i \cos \theta}{\cos \theta + i \sin\theta} \cdot \frac{\sin\theta }{\cos\theta} \, . $$ Finally convince yourself that the first factor is equal to $i$, and you are done.

score 1 · Answer 3 · answered Oct 15 '18 at 07:49

1

$$ \tan\theta=\frac{\sin\theta}{\cos\theta} =\frac{\dfrac{z-\bar{z}}{2i}}{\dfrac{z+\bar{z}}{2}}= \frac{1}{i}\frac{z-\bar{z}}{z+\bar{z}}=\frac{1}{i}\frac{z-z^{-1}}{z+z^{-1}}= \frac{1}{i}\frac{z^2-1}{z^2+1} $$

answered Oct 15 '18 at 07:49

egreg

238,574

score 1 · Answer 4 · answered Oct 15 '18 at 07:52

$\dfrac{z^2-1}{z^2+1}$ reminds me of Componendo and Dividendo?

$\dfrac{z^2}1=(\cos\theta+i\sin\theta)^2=\dfrac{\cos\theta+i\sin\theta}{\cos\theta-i\sin\theta}$ as $\dfrac1{\cos\theta-i\sin\theta}=\cos\theta+i\sin\theta$

Now apply Componendo and Dividendo

Complex number problem $i \tan (\theta)$ proof.

4 Answers4